Under ideal conditions, the velocity of steam at the outlet of a nozzle for a drop of 450 kJ/kg from inlet reservoir condition up to the exit is:

Concept:

The energy changes in a flow can be analyzed by steady flow energy equation (SFEE)

It is given as

\({h_1} + \frac{{V_1^2}}{2} + {Z_1} \times g + Q = {h_2} + \frac{{V_2^2}}{2} + {Z_2} \times g + W\)

Where h is the enthalpy of the flowing stream and V²/2 represents the kinetic energy and Zg represents the potential energy, Q and W ae the heat and work interaction respectively

Calculation:

Given,

Change in enthalpy h₁ - h₂ = 450 KJ/Kg

Now applying steady flow energy equation by the assumption (W = 0, Q = 0, ΔP.E = 0)

For ideal condition of nozzle

The initial velocity is assumed zero as it is very less compared to the exit velocity

\({h_1} = {h_2} + \frac{{V_2^2}}{2}\)

\({V_2} = \sqrt {2\left( {{{\rm{h}}_1} - {{\rm{h}}_2}} \right)} = \sqrt {2 \times 450 \times 1000} = 948.68\;m/sec\;\)